By "ck" I'm assuming you're referring to the series solution's coefficients, i.e. \(y=\displaystyle\sum_{k\ge0}c_kx^k\).
Differentiating gives
\(y'=\displaystyle\sum_{k\ge1}kc_kx^{k-1}\)
\(y''=\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}\)
Substituting into the ODE yields
\(\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}-x^2\sum_{k\ge1}kc_kx^{k-1}+x\sum_{k\ge0}c_kx^k=0\)
\(\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}-\sum_{k\ge1}kc_kx^{k+1}+\sum_{k\ge0}c_kx^{k+1}=0\)
By extracting the first term (\(k=0\)) of the third series, you can consolidate the second and third series:
\(\displaystyle\sum_{k\ge0}c_kx^{k+1}=\sum_{k\ge1}c_kx^{k+1}+c_0x\)
Next, since the first series starts with a constant term while the other two start with a quadratic term, remove the first two terms:
\(\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}=\sum_{k\ge4}k(k-1)c_kx^{k-2}+2c_2+6c_3x\)
Now shift the index so that the series begins at the same starting index by replacing \(k\) with \(k+3\).
\(\displaystyle\sum_{k\ge4}k(k-1)c_kx^{k-2}=\sum_{k\ge1}(k+3)(k+2)c_{k+3}x^{k+1}\)
So now the series can be joined into one:
\(\displaystyle\sum_{k\ge1}\bigg[(k+3)(k+2)c_{k+3}-kc_k+c_k\bigg]x^{k+1}\)
which means the ODE is equivalent to
\(\displaystyle\sum_{k\ge1}\bigg[(k+3)(k+2)c_{k+3}-(k-1)c_k\bigg]x^{k+1}+c_0x+2c_2+6c_3x=0\)
It follows that \(2c_2=0\implies c_2=0\) and \(c_0+6c_3=0\). Using this, you solve the recurrence
\((k+3)(k+2)c_{k+3}-(k-1)c_k=0\iff c_{k+3}=\dfrac{k-1}{(k+3)(k+2)}c_k\)
with the initial values \(c_0\) and \(c_1\).
Because \(c_2=0\), it follows that \(c_5=0,c_8=0,\ldots\); in general, \(c_{k=3\ell-1}=0\) for all \(\ell\ge1\).
Also, because the RHS of the recurrence vanishes when \(k=1\), it follows that \(c_4=0,c_7=0,\ldots\); or generally, \(c_{k=3\ell+1}=0\) for \(\ell\ge1\), leaving you with only one not-necessarily-zero term of \(c_1\).
So the hard part is finding a solution for \(c_k\) when \(k\) is a multiple of 3. By finding the first few of these terms, you'll start to notice a pattern. Starting with \(c_0\), you find
\(c_3=-\dfrac1{3\times2}c_0\)
\(c_6=\dfrac2{6\times5}c_3=-\dfrac1{6\times5\times3}c_0\)
\(c_9=\dfrac5{9\times8}c_6=-\dfrac1{9\times8\times6\times3}c_0\)
\(c_{12}=\dfrac9{12\times11}c_9=-\dfrac1{12\times11\times9\times6\times3}c_0\)
and so on, following a general pattern of
\(c_{k=3\ell}=-\dfrac{c_0}{(3\ell-1)\prod\limits_{i=1}^\ell 3i}\)
\(c_{k=3\ell}=-\dfrac{c_0}{3^\ell\ell!(3\ell-1)}\)
valid for \(\ell\ge1\).
So the series solution is
\(y=\displaystyle\sum_{k\ge0}c_kx^k\)
\(y=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+c_6x^6+\cdots\)
\(y=c_0+c_1x+\displaystyle\sum_{\ell\ge1}c_{3\ell-1}x^{3\ell-1}+\sum_{\ell\ge1}c_{3\ell}x^{3\ell}+\sum_{\ell\ge1}c_{3k+1}x^{3k+1}\)
\(y=c_0+c_1x+\displaystyle\sum_{\ell\ge1}c_{3\ell}x^{3\ell}\)
\(y=c_0+c_1x-c_0\displaystyle\sum_{\ell\ge1}\frac{x^{3\ell}}{3^\ell\ell!(3\ell-1)}\)
\(y=c_0+c_1x+c_0\displaystyle\sum_{\ell\ge1}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}\)
\(y=c_1x+c_0\displaystyle\sum_{\ell\ge0}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}\)
So the two solutions are \(y_1=x\) and \(y_2=\displaystyle\sum_{\ell\ge0}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}\).
To solve the IVP, notice that when \(x=0\), you have
\(y(0)=\displaystyle\sum_{k\ge0}c_kx^k=c_0=2\)
\(y'(0)=\displaystyle\sum_{k\ge1}kc_kx^{k-1}=c_1=5\)
so the solution to the IVP is the same as that to the general solution with the unknowns replaced accordingly.
